Question: Simplify and expand the following expression: $ \dfrac{3}{x + 1}+ \dfrac{5}{3x + 12}- \dfrac{4}{x^2 + 5x + 4} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the second term: $ \dfrac{5}{3x + 12} = \dfrac{5}{3(x + 4)}$ We can factor the quadratic in the third term: $ \dfrac{4}{x^2 + 5x + 4} = \dfrac{4}{(x + 1)(x + 4)}$ Now we have: $ \dfrac{3}{x + 1}+ \dfrac{5}{3(x + 4)}- \dfrac{4}{(x + 1)(x + 4)} $ The least common multiple of the denominators is: $ (x + 1)(x + 4)$ In order to get the first term over $(x + 1)(x + 4)$ , multiply by $\dfrac{3(x + 4)}{3(x + 4)}$ $ \dfrac{3}{x + 1} \times \dfrac{3(x + 4)}{3(x + 4)} = \dfrac{9(x + 4)}{(x + 1)(x + 4)} $ In order to get the second term over $(x + 1)(x + 4)$ , multiply by $\dfrac{x + 1}{x + 1}$ $ \dfrac{5}{3(x + 4)} \times \dfrac{x + 1}{x + 1} = \dfrac{5(x + 1)}{(x + 1)(x + 4)} $ In order to get the third term over $(x + 1)(x + 4)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{4}{(x + 1)(x + 4)} \times \dfrac{3}{3} = \dfrac{12}{(x + 1)(x + 4)} $ Now we have: $ \dfrac{9(x + 4)}{(x + 1)(x + 4)} + \dfrac{5(x + 1)}{(x + 1)(x + 4)} - \dfrac{12}{(x + 1)(x + 4)} $ $ = \dfrac{ 9(x + 4) + 5(x + 1) - 12} {(x + 1)(x + 4)} $ Expand: $ = \dfrac{9x + 36 + 5x + 5 - 12}{3x^2 + 15x + 12} $ $ = \dfrac{14x + 29}{3x^2 + 15x + 12}$